Question

Consider the reaction mg(s)+fe2+(aq)→mg2+(aq)+fe(s) at 79 ∘c , where [fe2+]= 3.50 m and [mg2+]= 0.310 m . part a what is the value for the reaction quotient, q, for the cell? express your answer numerically.

Answer 1

The following reaction is given:

$Mg_{(s)}+Fe^{2+}_{(aq)}\to Mg^{2+}_{(aq)}+Fe_{(s)}$

So, the reaction quotient is:

$Q=\dfrac{[Products]}{[Reactants]}\\\\ Q=\dfrac{[Mg^{2+}]}{[Fe^{2+}]}$

Note that in the reaction quotient formula we only put the substances in the aqueous state. The substances in the solid state are not considered.

Then, we'll use the values of the statement:

$Q=\dfrac{[Mg^{2+}]}{[Fe^{2+}]}=\dfrac{0.31~M}{3.50~M}\\\\ Q\approx0,08857\\\\ \boxed{Q\approx8,857\times10^{-2}}$